lim _{x rightarrow 0} frac{e^{2|sin x|}-2|sin | vedclass

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(D) Let $t = |\sin x|$. As $x ightarrow 0$,$t ightarrow 0^+$. The expression becomes $\lim _{t ightarrow 0^+} \frac{e^{2t}-2t-1}{t^2} 	imes \fr

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is equal to $2$

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(D) Let $t = |\sin x|$. As $x ightarrow 0$,$t ightarrow 0^+$. The expression becomes $\lim _{t ightarrow 0^+} \frac{e^{2t}-2t-1}{t^2} 	imes \frac{\sin^2 x}{x^2}$. Since $\lim _{x ightarrow 0} \frac{\sin^2 x}{x^2} = 1$,we evaluate $\lim _{t ightarrow 0^+} \frac{e^{2t}-2t-1}{t^2}$. Using the Taylor series expansion $e^{2t} = 1 + 2t + \frac{(2t)^2}{2!} + \frac{(2t)^3}{3!} + \dots = 1 + 2t + 2t^2 + \frac{4t^3}{3} + \dots$. $\lim _{t ightarrow 0^+} \frac{(1 + 2t + 2t^2 + \dots) - 2t - 1}{t^2} = \lim _{t ightarrow 0^+} \frac{2t^2 + \dots}{t^2} = 2$. Therefore,the limit is $2 	imes 1 = 2$.

$\lim _{x \rightarrow 0} \frac{e^{2|\sin x|}-2|\sin x|-1}{x^2}$

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